I P=U*I: Why do electrons drop ALL energy e*U in the circuit?

Consider a simple circuit consisting of a voltage source ##U## and a load with resistance ##R##, e.g. a lamp or a motor. The current is given by ##I=U/R##. The number of electrons passing the circuit per second is ##n=I/e##. The power consumed by the load is calculated by
$$P=U\cdot I=U\cdot e\cdot n=\Delta E\cdot n\enspace,$$
where ##\Delta E=U\cdot e## is the (kinetic) energy an electron would gain travelling from the negative to the positive pole of the power source if there was no load.

In this computation, we assume the electron gives all its energy to the load and has kinetic energy zero when it arrives at the plus pole of the power source. But why is that? Why can't the electron maybe only lose half its energy to the load and still have kinetic energy when it enters the battery?

Staff: Mentor

Staff: Mentor

Consider what happens when you turn on a light switch. The change in the fields moves at a little less than the speed of light. By contrast, the electrons move at about a mm/s. How quickly does the energy get from the source to the light? Is it something that happens nearly at the speed of light or closer to a mm/s?

Ok, then I think I need a different derivation of ##P=U\cdot I##. Because the equation I wrote in #1 starting from the right is basically the derivation I learnt in school, i.e. looking at the energy difference of an electron (or any other current carrier) between the poles of a voltage source and then counting how many of them pass the circuit per second.

How can ##P=U\cdot I## be derived alternatively without looking at individual charges and make assumptions about how much energy they drop in the circuit?